Question

Directions for Confidence Interval Problems: For Questions 1 to 3, include the following information in your solution: • Does your problem involve 1 population or 2 populations? . State a. • State whether you should use z or X2 • Find / provide the appropriate value (z or x2) from the table (or your calculator). Round z-values to 2 decimals, and round X2-values to 3 decimals. • Find the confidence interval. Round your final values to 3 decimals. • State if your confidence interval is for p, Pi-P2, oor o. • Write a verbal interpretation of your confidence interval. (For example: We are xx% confident....) • For questions that involve 2 populations write a conclusion as to whether pı > P2, P1 < p2, or P1 = P2. 3. A class has been given a statistics exam. In a random sample of 12 students, the test scores are 65, 73, 74, 68, 70, 74, 69, 71, 75 73 67 70, giving us the sample statistics i = 70.75 and s2 = 9.84. Assume a normal distribution of scores. (a) Construct a 98% confidence interval for o2. (b) Construct a 98% confidence interval for o. (Use your information from Part A, and just state your confidence interval.)

Answer 1

Solution We have been given the test scores foź statistics exam of Sandomly selected 12 students in a class and w The selected Zandom sample of 12 students gives the sample statistics as follows; ī = 70.75 and 5² = 9.84 where ] = Sample mean o sé a sample Variance We need to find the 98 %_confidence intervals for population Vaciance (02) and population standard deviation () fox a single population. - Here we use 2? distribution to find C.I. forę o? and o - Since confidence level (@= 0.98 the significance level (QC) = 1-C = 1-0.98 = 0.02 - To find confidence intervals for a² and so we require to find critical values of chi-square as follows, : & k-2, 1-1 and tu= larant where = significance level ne numbes obsezvations / sample size 2-1 = degrees of freedom for X-distribution Here we have, - a= 0.02 & 042= 0.01 and df = n-t = 12-1 of c11

1-4 = 0.99 d = 0.01 df = 11 From the_chi-square table_ive fet, - Ko.99,= 3.053 and - Lys Korotytt = 24.725 - The 100% (1-6) % confidence interval foc population variance o² is given by; . 0² . 11*9.84 7.24.725 11*9.89 - 3.053 4.378 < 02 < 35.454 .98% confidence interval foz oz is, (4.378,35:454) - This can be interpreted as We aze_98% confident that true parameter value for a lies within (4.378, 35.454) - - Similarly 100* (1-)% confidence interval foz population standard deviation r is given by;

114 9.84 24.725 _lo</ 11* 9.84 3.053 2.092 so < 5-954 .: 98% confidence interval faz o is, (2.092 ,5.954) - This can be interpeeted as; We are 98% confident that true parametes value foro lies within (20092, 5-954)